Problem: Roselyn is driving to visit her family, which live $150$ kilometers away. Her average speed is $60$ kilometers per hour. The car's tank has $20$ liters of fuel at the beginning of the drive, and its fuel efficiency is $6$ kilometers per liter. Fuel costs $0.60$ dollars per liter. How long can Roselyn drive before she runs out of fuel?
Explanation: There can be many ways to solve this problem. Here, we will do this by thinking about units. Let's say Roselyn can drive $x\,\text{hours}$ before she uses up $20\,\text{liters}$ of fuel. How can we relate these two quantities with an equation? $\begin{aligned} 20\,\text{liters}\cdot y\,\dfrac{\text{hours}}{\text{liter}}=x\,\text{hours} \end{aligned}$ So in order to find the duration $x$, we need to figure out the value of $y$, which is the rate of hours per liter. Notice what other information we are given: $150\,\text{kilometers}$ $60\,\dfrac{\text{kilometers}}{\text{hour}}$ $6\,\dfrac{\text{kilometers}}{\text{liter}}$ $0.6\,\dfrac{\text{dollars}}{\text{liter}}$ Which of these quantities can help us calculate a rate whose units are $\dfrac{\text{hours}}{\text{liter}}$ ? We can combine the following quantities: $\begin{aligned} &\phantom{=}\dfrac{6\,\dfrac{\text{kilometers}}{\text{liter}}}{60\,\dfrac{\text{kilometers}}{\text{hour}}} \\\\ &=\dfrac{6}{60}\,\dfrac{\cancel\text{kilometers}}{\text{liter}}\cdot\dfrac{\text{hours}}{\cancel\text{kilometer}} \\\\ &=0.1\,\dfrac{\text{hours}}{\text{liter}} \end{aligned}$ Now we can plug that in the original equation: $\begin{aligned} 20\,\text{liters}\cdot 0.1\,\dfrac{\text{hours}}{\text{liter}}&=x\,\text{hours} \\\\ 2\,\text{hours}&=x\,\text{hours} \end{aligned}$ In conclusion, Roselyn can drive $2$ hours before she runs out of fuel.